Point Set Topology - III | Mathematics for Competitive Exams PDF Download

Interior And Exterior Of A Set

Interior Points of A Set

A point x is an interior point of a set S is S is a nbd of x. In other words, x is an interior point of S if ∃ an open interval (a, b) containing x and contained in S, i.e., x ∈ (a, b) ⊆ S.

Thus a set is a neighbourhood of each of its interior points.

Interior of A Set : The set of all interior points of set is called the interior of the set. The interior of a set S is generally denoted by Sj or into S.

Example : The interior of the set N or 1 or Q is the null set, but interior of R is R.

Example : The interior of a set S is a subset of S, i.e., Sj c S.

Interior, Exterior and Boundary of A Set
Definition :

(i) Let A be a subset of R and let p e A. Then p is called an interior point of A is there exists an e-nbd of p contained in A, i.e., if there exists an c > 0 such that ] p - ε, p - ε [ ⊂ A. The set of all interior points of A is called the interior of A and is denoted by Aº or by int A.

(ii) A point p is called an Exterior point of A if there exists an e-nbd of p contained in the complement A’ of A. The set of all Exterior points of A is called Exterior of A and is denoted by Ext A.

(iii) A point p is called a boundary point (or frontier point) of a if it is neither an interior nor an Exterior nor an Exterior point of A. The set of all boundary points of A called the boundary (or frontier) of A and is denoted by b (A) [or Fr(A)].

Result :

(i) Aº is the set of all those points of A which are not the limit points of A.’

(ii) A point p is an Exterior point of if and p is not an adherent point of A.

(iii) A point p is a boundary point of A if and only if every ε-nbd of p intersects both A and A’.

(iv) Aº is an open set,

(v) Aº is the largest open set contained in A

(vi) A is open if and only if Aº = A.

Theorem : Let A, B be any subsets of R Then

(i) Aº ⊂ A,
(ii) A ⊂ B ⇒ Aº ⊂ B°

(iii) (A ∩ B)° = A° ∩ B°
(iv) A° ∪ B° ⊂ (A ∪ B)°

(v) A°° = A°

Proof : We prove only (iii). The proofs of others are easy and let as Exercise for the reader. We have by (ii)

A ∩ B ⊂ A ⇒ (A n B)° ⊂ A° and A ∩ B ⊂ B ⇒ (A ∩ B)° ⊂ B.

(A ∩ B)° ⊂ A° ∩ B°    ...(1)

Again A° ⊂ A and B° ⊂ B ⇒ A° ∩ B° ⊂ A ∩ B.

Also A° ∩ B° is open, being    the intersection of two open sets.

Thus A° ∩ B° is an open set contained in A ∩ B. But (A ∩ B)° is the largest open set contained in A n B. Hence

A° ∩ B° ⊂ (A ∩ B)°    ...(2)

From (1) and (2), (A ∩ B)° = A° ∩ B°

Compactness

(i) Open Cover of A Set : Definition : Let S ⊂ R and C = {A_iI i ∈ I}, R be the group of open set in R. Then C is called the open cover of the set S, if

i.e., every element of S exist atleast is one of the open set of C.
(ii) Finite Subcover : Definition : Any finite subcover C’ = {A_j I j ∈ J ⊂ I}, S of open subcover C = {A_i I i ∈ 1} of any set S is called a finite subcover of S, if

Example : Group C = {(- n, n) | n ∈ N} and C’ = {(- 3n, 3n) | n ∈ N} are two open covers of R and C’ is an open cover of C.
Example : Group of open intervals is the open cover of the set (0, 1).

Compact Set : Definition

A non-empty subset of R is said to be compact if a finite subcover of S is contained in each open cover of any set S.

Example : A finite subset S = {x,, x2,    xj of R is a compact set, since if C = {A_i I i ∈ 1} is an open cover of S, then for every x_r ∈ S, there exist an open Ar in C, such that x_r e A_r, r = 1, 2,.........n.

Now if C’ = {A I r = 1, 2    n}, then
Therefore C’ is a finite cover of the set S.

Consequently, S is compact set.

Heine Borel (H-B) Theorem : Statement

Every open cover of a closed and bounded subset S of real number contains a finite subcover of S.

Proof. Let S be a closed and bounded subset of the set R of real numbers.

Now since S is bounded, therefore there exists an closed interval
l₀ = [a, b] such that S ⊂ [a, b]

where a and b are the infimum and supremum respectively of S.

Let C = {A_i I i ∈ N} be an open cover of [a, b].

Hence C is also an open cover of S.
Therefore
If possible, suppose no finite subcover of [a, b] is contained in C i.e., any finite sub group of C is not the subcover of l₀ = [a, b].
Then bisecting  there will not exist finite subcover in C of atleast one of the intervals
Denote this interval by l₂ = [a₂, b₂].

Repeating this process, we get a sequence of nested intervals

I₀ ⊃ I₁ ⊃ I₂ ⊃..... ⊃ I ⇒

in which no finite cover of any interval exist in C.
Clearly, Length of I = II_nI = b_n - a_n =

Therefore by the property of nested intervals, there exist only one point say x₀ in This point x₀ exists in each of the above nested interval, therefore x₀ ∈ [a, b]. But C is the open cover of [a, b]. Therefore for x₀, there exists an open set say A₀ in C such that x₀ ∈ A₀. Now since A₀ is an open set, therefore x₀ ∈ A₀ ⇒ ∃ε > 0 s.t. (x₀ - ε x₀ + ε) ⊂ A₀.
Again, since lim II_nI = 0, therefore a natural number k may be chosen such that b then l_k ⇒ (x_o - ε, x₀ + ε) c A₀
Consequently, there will be a finite cover of one element of the interval l_k in C. This is contrary to the fundamental property of the above intervals l₀, I₁, l₂, ... .
Thus our assumption is false.

Therefore there exist a finite cover of [a, b] in C.

Hence finite cover of S is also in C.

The above theorem can also be stated as follows:
Alternative definition of Compact set
A non-empty subset of R is said to be compact if it is closed and bounded.

Properties of Compact Sets
Theorem : Every compact subset of real numbers is closed and bounded. A subset of real numbers is compact iff it is closed and bounded.

Proof. S is bounded :
Let S be the compact subset of R and A_n = (- n, n), then
C = {A_n I n ∈ N}

is an open cover of R. Hence this is also an open cover of S.

Since S is a compact set, therefore there exist finite natural numbers n₁, n₂, ..., n_k such that finite subgroup { An₁, A n₂,.. . . An_k } of C will also be subcover of S. Let n₀ = max {n₁, n₂...... n_k}, then clearly,

S ⊂ An₀ = ( - n₀,n₀)
Consequently, S is a bounded set.

S is closed :
For this we shall show that no point of R - S is limit point of S i.e. all the limit points of S exist in S itself.

Therefore let a ∈ R - S, then a ∉ S. For every n ∈ N,
is a closed set. ⇒ R - B_n is an open set.
⇒ C’ = (R - B_n I n ∈ N} is a group of open set.
Again, clearly  Now since a ∉ ( S , therefore
S ⊂ [R - ∩ B_n : n ⊂ N] = ∪ [R - B_n : n ∈ N] [By Demorgan’s law]
⇒ C’ is a cover of S
But S is a compact set, therefore by the compactness property, there will exist finite natural numbers m₁, m₂, .... m_s such that each point of S will exist atleast in one of

R - B_m1, R - B_m2 ,...., R - Bm_s
Therefore if x ∈ S, then
x ∈ S ⇒ 3 i ∈ {1, 2, 3, .... r} r.t. x ∈ R - Bm₁
⇒ any point of S does not exist in
⇒ a is not the limit point of S ⇒ all the limit points of S exist in S.
⇒ S is closed

Theorem. R is not compact.
Solution. Let An = (- n, n), then A_n is an open set for each value of natural number and C = {A_n I n ∈ N} is the open cover of R,
because
If C' = {An₁, An₂ ,......., An_k}
be the finite subgroup of the cover C, then let
n₀ = max {n₁, f n₂...., n_k}
then n0 ∉ A_n, ∀ i = 1, 2, .... k.
Therefore clearly, any finite subset of C can not be the cover of R.
Consequently, R is not compact.

Theorem : No open interval (a, b) is compact.
Solution. Let An then for every natural number n, A_n is an open set and C = {A_n I n ∈ N} is an open cover of (a, b) since (a, b) ⊂
Now we will show that no finite subgroup of C can not be the subcover (a, b). If
C' = { An₁, An₂ ,...., An_k}
be the finite subgroup of the cover C, then let n₀ = max { n₁, n₂, ..., n_k} then clearly,
Therefore C’ can not be the subcover of (a, b).
Consequently, (a, b) is not compact.

Cantor Ternary Set : [г]

I Stage : Denote the closed interval [0, 1] by F₀.

Divide the interval into three equal parts and remove the middle one third open interval l_1,1 =
Remaining two disjoint open intervals
Now J_1,1 U J_1,2 = F₁ (say)
II Stage : Now again divide the above two closed intervals J . . and J, 2 into three equal parts and remove the middle one third open intervals of each of
Thus there remains the following four (22) disjoint closed intervals :

The length of each of the above interval = 1/9
Now J_2,1 U J_2,2 U J_2,3 U J_2,4 = F₂ (say)   ...(2)
and O₂ = l_2,1 U l_2,2

n^th Stage :
In the nth stage (2^n-1) open intervals l_n,1, l_n,2,...,l_n,2ⁿ - 1 , will be removed and there remains 2ⁿ closed intervals l_n,1, l_n,2, I_n,2ⁿ (left to right) where length of each subinterval =


Now the setis called Cantor’s theory sat and is denoted by Γ'.
Remark. The set T is non-empty because the end points of the closed intervals have not been removed and are in the set T.

Connectedness

Connected Set. Definition : A subset S of the set of real numbers R is called connected set if no two disjoint open sets A and B of R exist such that

S ⊂ A ∪ B and A ∩ S ≠ ϕ, B ∩ S ≠ ϕ

The set which is not connected, is known as Disconnected Set.

Example 1 : S = {x ∈ R I 0 < x < 1} is a connected set because no two disjoint subsets A and B of R such that

S ⊂ A ∪ B and A ∩ S ≠ ϕ), B ∩ S ≠ ϕ

Example 2 : Each open interval (a, b) is a connected set.

Theorem : Any subset S of real numbers is a connected set iff S is an interval.

Proof. S is connected ⇒ is an interval :

If possible, let S be not an interval, then there will exist three real numbers a, q, b such that a < q < b and a g S. b ∈ S but q ∉ S.

Clearly, a ∈ (- ∞, q) = A (say), b e (q, ∞) = B (say)

Therefore there exist non-empty open sets A and B such that
A ∩ B = ϕ, S ⊂ A ∪ B, S ∩ A, ≠ 4. S ϕ. B = ϕ

Consequently, S is not a connected set.

Therefore, if S is not an interval, then this will not be a connected set.

Conversely : Set S is an interval ⇒ S is connected
Let S be an interval and A, B be two open sets such that
A ∩ S ≠ ϕ, B ∩ S ≠ ϕ and A ∩ B = ϕ
Let a ∈ A ∩ S and b ∈ S ∩ S. since a ≠ b,

Therefore, let a < b then [a, b] will be a subinterval of the set S.

Let u be the supremum of the set A ∩ [a, b].

Since a, b are internal points of A and B respectively
and A ∩ B = ϕ, therefore a < u < b

Now if ε > 0 is a number such that a < u - ε < u + ε < b
then there will be atleast one point of A in (u - ε, u)
and no point of A will be there in (u, u + ε).

Consequently, (u - ε, u + ε) is not the subset of the sets A or B.

Again, since A and B are open, therefore u ∉ A and u ∉ B.

S ⊄ A ∪ B

⇒ S is a connected set.

Therefore S is connected ⇒ S is an interval.

Theorem : The set R of real numbers is a connected set.

Proof. If possible, let R be not connected set then there exist two disjoint open sets A and B such that

R ⊂ A U B and A ∩ R ≠ ϕ, B ∩ R ≠ ϕ)

Let a ∈ A, b ∈ B and

A₁ ={x ∈ [0, 1] :a + x(b - a) ∈ A}
and B₁ = {x ∈ [0, 1] : a + x(b - a) ∈ B}

Now since [a, b] = {α = a + s(b - a) : 0 < x < 1} ⊂ R ⊂ A ∪ B
∴ A₁ ∪ B₁, = [0, 1]

Again since a ∈ A and a = a + 0(b - a) ⇒ 0 ∈ A, ⇒ A. ≠ ϕ

Similarly, b ∈ B and b = a + 1(b - a) ⇒ 1 ∈ B₁ ⇒ B₁ ≠ ϕ

Since A, B are two disjoint open sets, therefore At and Bt are also two disjoint open sets. Consequently {A,, B,} shows the disconnectivity of the interval [0, 1] which contradicts the statement of the theorem proved earlier.

Therefore our earlier assumption is false i.e., R is a connected set.

Example 32: That the only subsets of R which are both open and closed are R and ϕ.

Let A be a subset of R which is both closed as well as open.

Then ~ A = R - A = B (say) will also be open as well as closed.

If A ≠ ϕ and A ≠ R, then there exist two non empty disjoint subsets A and B such that
R ⊂ A ∪ B, A ∩ R ≠ ϕ, B ∩ R ≠ ϕ

Consequently, R will not be connected set which contradicts the above theorem.

Therefore A = ϕ or R.

i.e., ϕ and R are the only two subsets of R which are both open and closed.

Power Series

Definition 1: A series of the form
where a₀,a₁,a₂,.......are constant, is called a power series in x.
Definition 2 : A constant R > 0 such that Σa_n(x - x₀)ⁿ converges for I x - x₀l < R and diverges for I x - x₀ I > R is called the radius of convergence of the series.

Interval (domain) of Convergence
The interval l x l < R or - R < x < R is called the interval (domain) of convergence.

Examples
1. converges when I x I < 1 and diverges when I x I > 1.

Radius of Convergence
The radius of convergence is a number R such that the series
Σa_n(x - x₀)ⁿ
converges absolutely for lx - x₀I < R and diverges for lx - x₀l > R

Note that :

• If the series converges ONLY at x = x₀, R = 0.
• If the series converges for ALL values of x, R is said to be infinite

Formula for The Radius of Convergence

If the power series ΣXa_nxⁿ is such that a_n ≠ 0 for all n and then ΣXa_nxⁿ is convergent (absolutely convergent) for I x I < R and divergent for I x I > R.
OR

The radius of convergence R of a power series Σa_n xⁿ may be determined by the following

R = ∞, when lim I a_n l^1/n = 0
R = 0, when lim I a_n l^1/n = ∞

Example 33: Find the radius of convergence of

Hence the radius of convergence = 1.

Example 34: Find the radius of convergence of the series

∴

Hence the radius of convergence = e.

Ex. Find the radius of convergence and the exact interval of convergence of the following power series :
(i)
(ii)
Ans. (i)
∴
Thus the given series converges for I x I < 1 and diverges for I x I > 1. We now check at the end points x = 1, -1.

For x = 1, the nth term of the series is

However for x = -1, the given series is an alternating series for which ur+1 < u_ V n and lim un=0. Thus, by Leibnitz’s test, the series converges for x = -1. Hence the interval of

n-**

convergence is [-1, 1[.
(ii) Given
Let x² = y
then power series reduces to

We have







= 4

So the power series converges for I y I < 4 i.e. Ixl²<4 ⇒ l x l < 2 and diverges for I y I > 4. i.e. I x I² > 4. So the interval of convergence is I x I < 2 i.e. (- 2, 2). And Radius of convergence is 2.

Example 35: Prove that the power series

has unit radius of convergence.

(omitting the first term)

∴

Example 36: Find the domain of convergence of the series
(i)
(ii)

(i)

For x = 1, the given power series is  which is convergent.
For x = -1, the given power series is

which is convergent, by Leibnitz’s Test.

Hence the domain for convergence is [-1, 1].

(ii) The given power series is about the point x = 1. We have

The domain of convergence of the given power series is
-2 + 1 < x < 2 + 1 or -1< x < 3.
For x = 3, given power series is  which is convergent by Leibnitz’s Test.
Hence the domain of convergence of the given power series is -1 < x < 3.

Example 37: Determine the interval of convergence of
(i)
(ii)
(iii) 

(i) a_n is coefficient of xⁿ. Thus a_n = 1/(2n+1) is wrong. Actually a_2n+1 = 1/(2n+1) and a_n = 0 whenever n is even. Thus, we apply R =

Hence, R = 1.
The interval of convergence is (-1, 1).
For x = 1, the series becomes
which is convergent by Leibnitz’s test.
For x = -1, the series becomes
which is again convergent.
Hence the Exact interval of convergence is [-1, 1].
(ii)

∴ R = 1.

Since the given power series is about the point x = 1, the interval of convergence is -1 + 1 < x < 1 +1 or 0 < x < 2.
For x = 2, the given series is  which is convergent by Leibnitz’s Test.
Hence the exact interval of convergence is (0, 2].
(iii) The given power series is about the point x = -2.


R = e and the interval of convergence is (- 2 - e, -2 + e).

Theorems of Power Series

Theorem : If a power series Σa_n xⁿ converges for x = x₀, then

(i) it is absolutely convergent in the interval I x I < I x₀ I

(ii) it is uniformly convergent in the interval I x I < I x. I, where I x₁ I < I x₀ I.

Proof. (i) Since Σa_n x₀ⁿ converges, so

Consequently there exists a positive integer m such that
... (1)


The series on the R.H.S. converges for I x I < I x₀1 (being a geometric series with common ratio < 1), it follows by the comparison test that Σ I a_nxⁿ I is convergent for I x₀ I < I x₀ I.

Hence Σ a_n xⁿ is absolutely convergent for I x I < I x0 I.
(ii) Let  Then Σ M_n converges, since I x₁ I < I x₀ I (being a geometric series with common ratio < 1).
Using (1), we obtain

I a_nxⁿ I < M_n ∀ n > m, where Σ M_n converges.

By the Weierstrass M-test, Σa_nxⁿ is uniformly convergent in I x I < I x₁ I.

Note : If a power series whose radius of convergence is 1, then it converges absolutely (- 1, 1) and uniformly in [- k, k], | k | < 1.

Theorem : If a power series ΣXa_nxⁿ converges for I x I < R, then it converges uniformly on [- R + ε, R - ε], for each ε > 0.

Proof. Let x₀ = R - ε < R so that Σa_nx₀ⁿ converges absolutely.

For I x I < x₀, I a_nxⁿ I < I a_n I x₀ⁿ.

Since Σa_nx₀ⁿ converges absolutely, therefore, by Weierstrass’s M-test, the series Σa_nx₀ⁿ converges uniformly on

[-x₀, x₀] = [-R + ε, R - ε].

Hence the theorem.

Theorem : Suppose a power series Σa_nxⁿ converges for I x I < R and that f(x) = Σa_n xⁿ. Then

(i) f(x) is continuous in (—R, R).

(ii) Σa_nxⁿ can be integrated term by term on (—R, R) and the integrated series has the same radius of convergence as Σa_nxⁿ

(iii) Σa_nxⁿ can be differentiated term and term on (-R, R) and f(x) = Σn a_nx^n-1 I x I < R. Further the differential series has the same radius of convergence as Σa_nxⁿ.

Proof, (i) Since each term of the power series Σa_nxⁿ is continuous in (—R, R) and the series is uniformly convergent on (-R, R), so the sum f(x) of the given series is continuous in (-R, R)

(ii) Since each term of X a_nxⁿ is continuous in (-R, R) and Σa_nxⁿ converges uniformly to f(x) in (-R, R), so Σ a_n xⁿ can be integrated term by term in (-R, R) and the integrated series is  Further

which shows that ∑a_nxⁿ and have the same radius of convergence.

(iii) We know each term of Z anxn is continuous, has continuous derivative in (-R, R) and ∑a_nxⁿ is uniformly convergent in (-R, R). Thus the series obtained by differentiating the power series ∑a_nxⁿ is ∑a_nx^n-1 Further

Hence the differentiated series Σn a_nx^n-1 is also a power series which has the same radius of convergence as Ian Σa_nxⁿ.

Note. If f(x) = Σa_nxⁿ, then f(x) is called the sum function of the series.

Abel’s Theorem : If a power series converges at an end point of its interval of convergence, then the power series is uniformly convergent in the interval which includes this end point.

Remarks : If a power series with interval of convergence (-R, R) converges at both the endpoints - R and R, then the power series is uniformly convergent in [- R, R].

Theorem : If a power series converges at the end-point x = R of the interval of convergence (-R, R), then it is uniformly convergent in (-R, R] or [-R + e, R] where c > 0. In particular, the series is uniformly convergent in [0, R],

Abel’s Limit Theorem : Let Σa_n xⁿ be a power series with finite radius of convergence R such that f(x) = Ea_n xⁿ - R < x < R. If the series Ea_n Rⁿ converges, then

Theorem : (Cauchy’s product of two power series)
Two power series can be multiple to obtain the power series where
c_n = a₀ b_n + a₁ b_n-1 + a₂ b_n-2 , +    + a_n b_n

the result being valid for each x within the common interval of convergence.

Example 38: Show that the series Exⁿ/n² is uniformly convergent in [-1, 1].

We have

The radius of convergence is R = 1 and so the given series is uniformly convergent in (-1, 1).
For x = 1, given series is which is convergent.
For x = -1, given series becomes

This is an alternating series for which u_{n + 1}, < u_n ∀ n and

So by Leibnitz’s test, the series to convergent for x = -1. Hence by Abel’s Theorem, given power series is uniformly convergent in [-1, 1].

Example 39: Show that the series

is uniformly convergent in [-1, k], 0 < k < 1.

We have

The radius of convergence is R = 1 and so the given series is absolutely convergent in ]-1, 1[.
For x = 1, given series  which is divergent.
However for x = 1, given series becomes  which is convergent by Leibnitz’s Test. Hence by Abel’s Theorem, given power series is uniformly convergent [-1, 1[ or [-1, k], where 0 < k < 1.

Example 40: Show that the following series

is uniformly convergent in [-1, 1].

Ignoring the first term, we have

∴
Since the radius of convergence is R = 1, given power series is absolutely convergent in (-1, 1). For x = 1, we see that the D’Alembert's ratio test fails. Now

By Raabe’s test, the series is convergent for x = 1.
Again for x = -1, the series is convergent.
(∵ ∑a_n is convergent ⇒ ∑a_n is convergent)
Hence the given power series is uniformly convergent in [-1, 1] (Abel’s Theorem).

Example 41: Prove that is uniformly convergent in [-1, 1).

Both the series given on the R.H.S. of (1) and (2) absolutely convergent in (-1, 1), so their Cauchy’s product converge to (1 - x)^-1 log (1 - x). Thus

Integrating, we get

where the constant of integration vanishes.
Since the series on the right converges at x = -1, so by Abel’s theorem, the above Expansion is valid for -1 < x < 1.

Example 42: Show that
(i)
(ii)
(iii) 

We have
(1 + x²)^-1 = 1 - x² + x⁴ - x⁶ + I x I < 1.    ...(i)
Radius of convergence is

Thus the series is convergent absolutely for all x such that I x I < 1, and is uniformly convergent in [-k, k], | k | < 1.

The integrated series is also convergent absolutely in (-1, 0) and uniformly in [-k, k], | k | < 1. Integrating (1), we get

where C is constant of integration.
Putting x = 0, we get C = 0 and so

Now the series
 is convergent at x = ±1.
is an alternating series which converges, by Leibnitz’s test.
Similarly it is convergent for x = -1].
By Abel’s theorem, it is uniformly convergent in [-1, 1].
Hence
(ii) Putting x = 1, by Abel’s limit theorem,

(iii) We have

and (1 + X²)^-1 = 1 - x² + x⁴ - x⁶ +..., -1 < x < 1.
By part (i), both the series are absolutely convergent in (-1, 1), so they can be multiplied and their Cauchy’s product will be equal to

integrating, we get

Here is the constant of integration vanishes on taking x = 0.

By Leibnitz’s test, the series on the R.H.S. converges for x = 1. Hence the above expansion is valid for -1 < x < 1.

Example 43: Show that

Deduce that

We know that

The series on the R.H.S. is a power series with radius of convergence 1 and so it is absolutely convergent in(-1, 1) and uniformly convergent for [-k, k], | k | < 1. The integrated series is also convergent in (-1, 1) and uniformly convergent for [-k , k], | k | < 1. integrating, we get

Putting x = 0, we get c = 0.
...(1)
For x = 1, the nth term of the series on the R.H.S. of (1) is


∴

So by Raabe’s test, the series is convergent for x = 1. Hence

(ii) At x = 1, by Abel’s limit theorem

Ex. Prove that

Example 44: For n > 0, we have

(1 + tⁿ)^-1 = 1 - tⁿ + t²ⁿ - t³ⁿ +....

It is clear that the series on the right has its radius of convergence R = 1.

By Theorem this series can be integrated term by term on (-1, 1) and the integrated series will also have the radius of convergence as unity. Thus
...(1)
At x = 1, the series on the right is convergent, by Leibnitz’s test.
Hence (1) is true for -1 < x < 1.

Theorem (Differentiation of Power Series)
Let a power series

have non-zero radius of convergence R and

Then, the following holds :
The function f is differentiable on the interval (c - R, c + R). Further the series  also has radius of convergence R and f(x) =
The function f has derivatives of all orders and

Theorem (Integration of Power Series)
Let

be a power series with non-zero radius of convergence R and let

Then
(i) The function f has an anti derivative F(x) given by

where C is an arbitrary constant and the series on the right hand side has radius of convergence R.
(ii) For [α, β] ⊂ (c - R, c + R),

where the series on the right hand side is absolutely convergent.

Example : Consider the power series

By the ratio test, for every x

Hence, the series is absolutely convergent for every x. Let

Then, f is differentiable by theorem and

Some Important Power Series

Consider the following power series :



For the first series,

Theorem : Define cosine and sine functions as sums of power series. Prove that

(i) S(x + y) = S(x)C(y) + C(x)S(y),

(ii) C(x + y) = C(x)C(y) - S(x)S(y),

where C, S denote cosine and sine respectively.

Proof. The cosine function is defined as
...(1)
The sine function is defined as
...(2)
These two power series are uniformly convergent ∀ x ∈ R and consequently, the functions C(x) and S(x) are continuous ∀ x ∈ R

Differentiating term by term the convergent series (2),

which gives us the uniformly convergent series (1),

Thus S(x) is derivable,  and S’(x) = C(x).    ...(3)

Similarly C’(x) = -S(x).    ...(4)

From (1) and (2); S(0) = 0 and C(0) = 1.    ...(5)

Let y be any arbitrary but fixed real number. We write
f(x) = S(x + y) - S(x)C(y) - C(x)S(y),    ...(6)
g(x) = C(x + y) - C(x)C(y) + S(x)S(y).    ...(7)
Differentiating w.r.t. x, we obtain
f(x) = C(X + y) - C(x)C(y) + S(x)S(y) = g(x).

g’(x) = -S(x + y) + S(x)C(y) + C(x)S(y) = -f(x).    [Using (3), (4)]

= 2f(x)g(x) - 2g(x)f(x) = 0 ∀ x ∈ R
It follows that f²(x) + g²(x) is a constant ∀ x ∈ R
⇒ f₂(x) + g²(x) = f²(0) + g2(0) ∀ x ∈ R

= [S(y) - S(0)C(y) - C(0)S(y)]² + [C(y) - C(0)C(y) + S(0)S(y)]² = 0    [Using (5)],
f²(x) + g²(x) = 0, ∀ x ∈ R implies that f(x) = 0 and g(x) = 0.
Hence (6) and (7) yield
S(x + y) = S(x)C(y) + C(x)S(y),
C(x + y) = C(x)C(y) - C(x)S(y).

Corollary 1. S(-x) = -S(x) ; C(-x) = C(x) ∀ x ∈ R
We have

Similarly we can prove the second part.

Corollary 2. S(x - y) = S(x)C(y) - C(x)S(y),
C(x - y) = C(x)C(y) + S(x)S(y).
Replacing y by - y in (i) and (ii) of Theorem

S(x - y) = S(x)C(-y) + C(x)S(-y),
C(x - y) = C(x)C(-y) - S(x)S(-y).
The desired results follow by Corollary 1.

Corollary 3. C²(x) + S²(x) = 1 or cos²x + sin² x = 1 ∀ x ∈ R
We have 1 = C(0) = C(x - x)

= C(x)C(-x) - S(x)S(-x)

= C²(x) + S²(x), by Corollary 1
Hence cos² x + sin² x = 1 ∀ x ∈ R

Corollary 4. I C(x) I < 1, I S(x) I < 1. ∀ x ∈ R

I cos x I < 1, I sin2 x I < 1 ; ∀ x ∈ R
This follows by Corollary 3.

Corollary 5. sin 2x = 2 sin x cos x, cos ²x = cos² x - sin² x.

Replacing y by x in (i) and (ii) of Theorem we obtain

S(2x) = 2S(x) C(x), C(2x) = C²(x) - S²(x)

Hence the results follows.

Remark. It may be noted that the above properties of C(x) and S(x) are similar to those of the trigonometric functions cos x and sin x respectively. Thus we have

C(x) = cos x and S(x) = sin x.

Theorem : Define cos x as a power series and determine its domain. Prove that there exists a least positive number π such that cos

Proof. cos x is defined as the sum of the power series

Its radius of convergence is

Thus this power series is convergent for ∀ x ∈ R and so the domain of cos x is R.
...(1)
∴ cos 0 = 1 > 0 and


Since the brackets are all positive, therefore

Thus we see that cos x is a continuous function in the closed interval [0, 2], where cos 0 and cos 2 are of opposite signs. By Intermediate value theorem, there exist at least one number α ∈ ]0, 2[ such that cos α = 0. We now proceed to show that α is unique.
Let, if possible, β ∈ (0, 2) satisfy cos β = 0 so that
cos α = cos β = 0, 0 < α, β < 2.
By Rolle’s theorem, there exists some number λ, α < λ < β, such that - sin λ = 0.

∴ sin λ = 0, 0 < λ < 2.    ...(2)

By definition,

∴
This contradicts (2).

Thus there exists one and only one root of the equation cos x = 0 lying between 0 and 2. i.e., cos α = 0, 0 < α < 2.
We denote α as Henceis the least positive root of the equation cos x = 0 i.e., cos
Also, therefore, we have cos x > 0 when 0 < x <
Theorem : Define Exponential function E(x) as the sum of a power series. Show that the domain is the set of all real numbers. Prove that
E(x + y) = E(x)E(y) for all x, y ∈ R.
If e denote E(1), prove that E(x) = E^x for all real x.

Proof. The Exponential function E(x) is defined as the sum of the power series
...(1)
We have
Thus the radius of convergence of the above power series is R = oo so that the power series is everywhere convergent i.e., E(x) is defined ∀ x ∈ R. This shows that the domain of the Exponential function is R.

It is clear that the function E(x), defined by (1), is continuous and has derivatives of all orders, for each x ∈ R.
Differentiating (1), we get

E’(x) = E(x) and so E”(x) = E(x), E’”(x) = E(x) etc.

in fact E⁽ⁿ⁾(x) = E(x)    ∀ n ∈ N.

By Taylor’s Theorem, for all real values t and x

or
Replacing t by x + y, we get

Thus E(x + y) = E(x) E(y) ∀ x, y ∈ R.    ....(2)
From (1),
The above series on the right converges to a number which lies between 2 and 3. We denote this number by e so that E(1) = e.
The property (2) can be Extended as
E(x₁ + x₂ +... + x_n) = E(x₁) E(x₂) ... E(x_n)    ...(3)

for any positive integer n and ∀ x ∈ R.

Taking x₁ = x₂ = ... = x_n = x in (3), we obtain

E(nx) = {E(x)}ⁿ ∀ x ∈ R, ∀ n ∈ N.    ...(4)

For x = 1, we obtain

E(n) = eⁿ ∀ n ∈ N    (∵ E(1) = e)    ...(5)

Taking x = m/n (m and n being positive integers) in (4),

E(m) = {E(m/n)}ⁿ or e^m = {E(m/n)}ⁿ, by (5)

E(m/n) = (e^m)^1/n = e^m/n.

Thus E(q) = e^q for all positive rationals q.    ...(6)

Let x be any positive irrational number. Then there always exists a sequence {x_n} of positive rational numbers such that x_n → x.
Using (6), E(x_n) = e^xn ∀ n ∈ N.

Since x_n → x as n → ∞ and E is a continuous function ∀ x ∈ R, therefore

E(x) = e^x for all positive irrationals.    ...(7)

From (6) and (7), we conclude that

E(x) = E^x for all positive reals.    ...(8)

Taking y = - x in (2), we get

E(0) = E(x)E(-x) or 1 = E(x) E(-x) (x > 0)    [∵ E(0) = 1, by (1)]
or
Hence E(x) = E^x ∀ x ∈ R.

Theorem : Let f be a continuous function on R such that

f(x + y) = f(x) + f(y) ∀ x, y ∈ R.    ...(1)

Show that f(x) = cx for some constant c and ∀ x ∈ R.

Proof. Putting y = x in (1), f(2x) = 2f(x).

In fact f(nx) = nf(x) ∀ x ∈ R, ∀ n ∈ N.

Replacing x by x/n, we get

f(x) = nf(x/n) or f(x/n) = (1/n) f(x) ∀ x ∈ R.

For any integers p and p(q > 0), it follows that

Thus f(rx) = r f(x), for any rational number r.    ...(2)

Let us write f(1) = e. Taking x = 1 in (2), we get

f(x) = cr, for every rational number r.    ...(3)

Let a be any irrational number. Then there exists a sequence {r_n} of rational numbers such that r_n → a

=> f(r_n) → f(a), by continuity of f

=> f(a) = lim f(r_n) = lim cr_n = c lim r_n = ca, using (3).

Hence f(x) = cx ∀ x ∈ R.

Theorem : Let f be a non-zero continuous function on R such that

f(x + y) = f(x) f(y) ∀ x, y ∈ R.    ...(1)

Show that f(x) = e^cx, for some constant c ≠ 0.

Proof. It is given that f(x) ≠ 0 ∀ x ∈ R. From (1),

Let g(x) = log_e f(x) ∀ x ∈ R. Then by (1),

g(x + y) = g(x) + g(y) ∀ x, y ∈ r.    ...(2)
Now g, being the composition of two continuous functions f(x) and log_e x, is itself continuous on R and g satisfies (2).
By Theorem, it follows that
g(x) = cx ∀ x ∈ R, c being a constant
⇒ log_ef(x) = cx    f(x) = e^cx ∀ x ∈ R.

Theorem : Let f be a non-zero function continuous on R⁺ such that f(xy) = f(x) + f(y) ∀ x, y > 0.    ...(1)

Show that f(x) = log_a x, x ∈ R+ (a > 0, a ≠ 1).

Proof. For x > 0, let us take t = loge x o x = e', t e R    ...(2)

Let g(t) = f(e') V t e R. Then g is continuous on R and

g(t + s) = f(e,+s) = f(e:es) + f(es)    by (1)

or    g(t + s) = g(t) + g(s).

Here t, s e R. By Theorem g(t) = ct, V t e R, c being a constant.

Thus f(x) = c loge x V x e R+, where c ≠ 0 as f is non-zero.

Let a = e1/s. Then f(x) = loga x V x e R+ (a > 0, a ≠ 1).

Taylor Series and Maclaurin Series

Definition : Let f : I = (a - 5, a + 8) -> R be a function which has derivation of all order in I. Then the power series

is called the Taylor series for f at x = a. We say f has Taylor series Expansion at x = a, if its Taylor series is convergent for x e I and its sum is f(x). For a = 0, the Taylor series for f is called the Maclaurin Series for f at x = 0.

Example : For the function its derivatives of all order exist in domain
I = (- ∞, 0) U (0, ∞)

For a = 1, since

f⁽ⁿ⁾ (x) = (- 1)ⁿ n!x^{- (n + 1)}
we have

f⁽ⁿ⁾ (1) = (- 1)ⁿ n!, ∀ n > 1
Thus 

n-0

is its Taylor series at x = 1. Since it is a geometric series, it will be convergent if I x - 1 | < 1 i.e.,    0 < x < 2.

Further, its sum is

Hence,  has Taylor series Expansion